9
$\begingroup$

My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:

$$|f(x)-f(y)|\geq k|x-y|$$

Then $f$ is bijective and its inverse is continuous.

Well, there's a Theorem , Invariance of domain, that says

"If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".

But I'm not knowing how to proceed...need a clue...Thanks for attention!!!

$\endgroup$
  • $\begingroup$ Invariance of domain in one dimension is not very hard to prove, it's essentially the statement that a strictly monotonic function on an interval has a continuous inverse which is what Brian proves in his answer. The case $n \geq 2$ is a completely different story... $\endgroup$ – t.b. Sep 16 '12 at 3:32
  • $\begingroup$ Related : math.stackexchange.com/questions/2825195 $\endgroup$ – Watson Jun 20 '18 at 14:56
10
$\begingroup$

You certainly don’t need invariance of domain.

Proving that $f$ is injective is easy: if $f(x)=f(y)$, then $|f(x)-f(y)|=0$, and therefore ... ?

Once you’ve shown that $f$ is injective, consider $f(0)$ and $f(1)$; they can’t be equal, so either $f(0)<f(1)$, or $f(0)>f(1)$. Show that in the first case $f$ must be strictly monotonically increasing, and in the second case it must be strictly monotonically decreasing; you’ll need the fact that $f$ is continuous.

After you’ve shown that $f$ is strictly monotonic, you can use the fact that $|f(x)-f(y)|\ge k|x-y|$ for all $x,y\in\Bbb R$ to show that the range of $f$ is unbounded in both directions. Then you can use continuity again to show that $f$ is surjective and hence a bijection.

At this point it should be easy to show that if $I$ is an open interval in $\Bbb R$, then $f[I]$ is also an open interval, which immediately implies that $f^{-1}$ is continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.